# Torque Practice: Angles, Lever Arms, and Turning Direction
*2026-11-03*
A 15 N force applied 0.80 m from a pivot can produce 12 N·m, 6 N·m, or no torque at all. Keep the pivot and contact point in place, then turn the force arrow. The distance to the contact point hasn't changed. The perpendicular lever arm has.
These eight original torque practice problems start with that change and build toward signed net torque and equilibrium. Sketch each setup and try the questions before scrolling to the worked answers. The numbers are simple enough that you can spend most of your attention on the geometry.
## Find the distance that belongs in the equation
For a force acting in the plane of the page, use:
```text
Torque magnitude: |τ| = rF sin θ = Fd_perp
```
Here `r` is the distance from the pivot to the force's application point. The torque angle `θ` is the angle from 0° to 180° between the pivot-to-application vector and the force vector. Compare the arrows' directions as though their tails were together. The **perpendicular lever arm**, `d_perp = r sin θ`, is the shortest distance from the pivot to the force's line of action. Extend the force arrow into a straight line to find it. These definitions follow [OpenStax's torque section](https://openstax.org/books/university-physics-volume-1/pages/10-6-torque).
Calculate the magnitude, then assign a sign: **counterclockwise positive, clockwise negative**, as viewed on the page. A positive torque describes a turning tendency; it doesn't establish the object's current direction of motion. Add signed torques about the same pivot to obtain net torque.
On each sketch, mark the reference point O, the force's application point, and its line of action. Label `r` and `d_perp` separately, then assign the torque sign. If you use an angle, mark which two directions form it. Torque uses `N·m`; the [SI units practice guide](/blog/si-base-and-derived-units/) explains why matching dimensions don't make torque and energy the same quantity.
## Eight problems to draw before calculating
Use ideal rigid rods or bars. All stated positions and force arrows lie in the page, with right as positive `x` and up as positive `y`. “Up” means up the drawing. The first seven problems have a fixed, frictionless pin at O. The pin can exert a force, but it exerts no torque about O. Neglect rod weight except in problem 7 and ignore other unspecified external loads. Each question describes the geometry at the instant considered. Treat the stated measurements as exact for these exercises; approximate decimal answers are rounded for readability.
### 1. Start with a perpendicular push
A horizontal rod extends rightward from O. A force of 15 N acts upward at P, 0.80 m right of O.
Find the perpendicular lever arm and the signed torque about O. State the turning direction.
### 2. Rotate only the force
Keep O, P, and the 15 N force magnitude from problem 1. The force now points up and right, **30° above the rightward rod**.
Find `θ`, `d_perp`, and the signed torque. What fraction of problem 1's torque remains?
### 3. Catch the complementary angle
Use the same O and P again. The 15 N force now points down and right. Its arrow makes **30° with the downward vertical**, toward the right.
Find `θ`, the angle between the rightward pivot-to-P vector and the force. Then calculate `d_perp` and the signed torque. Would using `sin 30°` give the correct magnitude?
### 4. Aim through the pivot
Keep P 0.80 m right of O. The 15 N force at P points horizontally left, directly toward O.
Find `r`, `θ`, `d_perp`, and torque about O. Explain how a nonzero force applied away from the pivot can give this answer.
### 5. Separate the radius from the lever arm
A bent rigid rod is pinned at O. A force acts at P, located 0.30 m right and 0.40 m above O. The force is **20 N horizontally rightward**.
Find the straight-line distance `r`, the perpendicular lever arm, and the signed torque about O. A student multiplies `(0.50 m)(20 N)`. Which distance did they use incorrectly?
### 6. Add torques on both sides
A horizontal bar extends to both sides of O. Three forces act on it:
| Application point relative to O | Force |
| --- | --- |
| 0.60 m right | 10 N upward |
| 0.40 m left | 15 N upward |
| 0.30 m right | 8 N downward |
Find each signed torque and their sum about O. Do the two upward forces turn the bar the same way?
### 7. Balance the beam, then check the forces
A uniform horizontal beam is 2.00 m long and weighs 40 N. Its left end is pinned at O. A 30 N load hangs 1.60 m right of O. A vertical cable at the beam's right end pulls upward. The beam and load remain at rest.
Draw the forces acting on the beam. Find the cable tension and the vertical force the pin exerts on it, using both a torque equation about O and a vertical force equation. The beam's own weight acts at its center.
### 8. Test the zero-net-torque claim
This time there is no pin. A uniform 2.0 kg bar is free to move in the plane on a frictionless horizontal surface. Gravity and the surface's support force balance perpendicular to that plane. View the bar from above; its center of mass is at O at the instant considered. It extends 0.50 m left and 0.50 m right of O.
A 10 N force points up the page at each end. These are the only in-plane forces. The bar starts at rest.
Find the net torque about its center of mass, the net in-plane force, and the center-of-mass acceleration. A student says, “Zero net torque means the bar stays at rest.” Is that conclusion justified? Here O is a reference point at the center of mass, **not a physical support**.
## Worked answers and geometry checks
### 1. The full distance is perpendicular
**`d_perp = 0.80 m`; `τ = +12 N·m`, counterclockwise.**
The pivot-to-P vector points right and the force points up, so `θ = 90°`:
```text
|τ| = (0.80 m)(15 N) sin 90° = 12 N·m
```
Lifting the right-hand end tends to turn the rod counterclockwise. In this case, `r` and `d_perp` happen to match.
### 2. Same contact point, half the torque
**`θ = 30°`; `d_perp = 0.40 m`; `τ = +6.0 N·m`.**
```text
d_perp = (0.80 m) sin 30° = 0.40 m
|τ| = (15 N)(0.40 m) = 6.0 N·m
```
The upward component still tends to lift the right-hand end. The rightward component acts along the rod and contributes no torque about O. The result is `6/12 = 1/2` of problem 1's torque, even though `r` and `F` are unchanged.
### 3. The marked 30° isn't θ
**`θ = 60°`; `d_perp ≈ 0.693 m`; `τ ≈ −10.4 N·m`, clockwise.**
Rightward and downward are 90° apart. The force lies 30° toward the right from downward, so the angle between the force and the rightward pivot-to-P vector is `90° − 30° = 60°`.
```text
d_perp = (0.80 m) sin 60° ≈ 0.693 m
|τ| = (0.80 m)(15 N) sin 60° ≈ 10.4 N·m
```
The downward component pushes the right-hand end down, giving a negative torque. Using `sin 30°` here would give a magnitude of 6.0 N·m and underestimate this torque. You could use `cos 30°` instead because this problem marks the angle from the perpendicular direction: the downward force component is `15 cos 30° N`.
### 4. The line of action passes through O
**`r = 0.80 m`; `θ = 180°`; `d_perp = 0`; `τ = 0 N·m`.**
The position vector points right while the force points left. Extend the force's horizontal line: it passes through O, so its perpendicular distance from O is zero.
```text
|τ| = (0.80 m)(15 N) sin 180° = 0 N·m
```
The force is nonzero, and its contact point is away from O. Its line of action still passes through O, so there is no turning tendency about O. A force applied at O itself would also have zero torque about O. Zero torque doesn't mean the force disappears from a force balance.
### 5. Use the distance to the horizontal force line
**`r = 0.50 m`; `d_perp = 0.40 m`; `τ = −8.0 N·m`.**
The pivot-to-P distance comes from the right triangle:
```text
r = √[(0.30 m)² + (0.40 m)²] = 0.50 m
```
The force's line of action is horizontal and 0.40 m above O. Its shortest distance from O is therefore 0.40 m. Pushing right above the pivot tends to turn clockwise:
```text
τ = −(20 N)(0.40 m) = −8.0 N·m
```
Multiplying `(0.50 m)(20 N) = 10 N·m` treats `r` as though it were perpendicular to the force. It isn't. If you know force components, an optional check is `τ = xF_y − yF_x`: `(0.30)(0) − (0.40)(20) = −8.0 N·m`.
### 6. Upward forces can oppose each other
**The torques are `+6.0`, `−6.0`, and `−2.4 N·m`; net torque is `−2.4 N·m`, clockwise.**
| Force | Turning tendency about O | Signed torque |
| --- | --- | --- |
| 10 N up, 0.60 m right | Counterclockwise | `+(0.60)(10) = +6.0 N·m` |
| 15 N up, 0.40 m left | Clockwise | `−(0.40)(15) = −6.0 N·m` |
| 8 N down, 0.30 m right | Clockwise | `−(0.30)(8) = −2.4 N·m` |
```text
τ_net = +6.0 − 6.0 − 2.4 = −2.4 N·m
```
Lifting the left end turns the bar clockwise; lifting the right end turns it counterclockwise. Assign the sign from the application point and force together. “Upward is positive torque” fails this check.
### 7. Torque balance gives 44 N; force balance gives 26 N
**Cable tension: 44 N upward. Pin force: 26 N upward.**
The beam's weight acts 1.00 m right of O. Both downward weights give clockwise torque, while the cable gives counterclockwise torque. The pin force acts at O and contributes zero torque about O:
```text
(2.00 m)T − (1.00 m)(40 N) − (1.60 m)(30 N) = 0
2.00T = 40 + 48
T = 44 N
```
Now include the pin force `R_y` in the vertical force balance:
```text
R_y + 44 N − 40 N − 30 N = 0
R_y = 26 N
```
The cable supports only part of the total 70 N load. The pin supports the rest. Leaving the pin force out of the torque equation is valid because its lever arm is zero; leaving it out of the force equation would be a mistake. There are no horizontal applied forces here, so the pin's horizontal force is zero.
### 8. Zero torque, but a 20 N net force
**`τ_net = 0 N·m`; net force is 20 N up the page; center-of-mass acceleration is 10 m/s² up the page.**
The left-end force gives clockwise torque and the right-end force gives counterclockwise torque:
```text
τ_net = −(0.50 m)(10 N) + (0.50 m)(10 N) = 0 N·m
F_net,y = 10 N + 10 N = 20 N
a_CM,y = 20 N / 2.0 kg = 10 m/s²
```
At this instant, the bar has zero angular acceleration about its center of mass and a nonzero translational acceleration. Because it starts at rest, it begins translating without starting to rotate. The student's conclusion fails the force check.
Equilibrium requires both zero net external force and zero net external torque. Static equilibrium also requires rest in the chosen inertial frame; see [OpenStax's equilibrium conditions](https://openstax.org/books/university-physics-volume-1/pages/12-1-conditions-for-static-equilibrium). Even for a bar on a fixed axle, zero net torque doesn't prove zero angular velocity: it may already be rotating at constant angular velocity. A torque calculation alone doesn't supply the initial motion.
If you're mixing up balanced forces with force pairs on different objects, use the separate [Newton's third law worksheet](/blog/newtons-third-law-practice/).
## Keep the geometry decision, then try fresh numbers
After checking your work, close the answers and redraw one missed setup. Change the force direction or move its application point to the other side of O. Predict the sign before calculating.
For repeated mistakes, a small recall prompt is useful:
| Front | Back |
| --- | --- |
| Which angle belongs in `rF sin θ`? | The angle between the pivot-to-application vector and the force vector. Check what the drawing's marked angle measures. |
| A force's line of action passes through O. What is its torque about O? | Zero: its perpendicular lever arm about O is zero. |
| Does zero net torque establish static equilibrium? | No. Check net force and whether the object is at rest too. |
The [AP Physics 1 Flashcards deck](/catalog/packages/ap-physics-1-flashcards/) has 400 English cards, including lever arms, torque signs, and rotational-motion conditions. It's a broader course review, not this worksheet's answer key. Nibomo is our product; the deck is an optional study resource hosted here.
Answering cards from memory and revisiting them in spaced reviews can help retain these recurring distinctions over time. Keep solving complete problems too, so you have to find the angle and lever arm in unfamiliar drawings. The [algebra-based physics study guide](/blog/how-to-use-flashcards-for-algebra-based-physics-1/) gives more examples of combining short recall with written practice.
*AI tools assisted substantially with source research, writing, editing, and construction of these original exercises. Numerical answers were checked with Python under the stated ideal models. The article has not been reviewed by a human physics expert. The illustration is AI-generated.*
**AP Physics 1 Flashcards: Complete 8\-Unit Course Review**
A 400\-card AP Physics 1 review of concepts, formulas, graphs, experiments, and reasoning across all eight course units\.
400 cards · Audience languages: English
[Study this deck for free]()
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